Proving $\mathbb{R}^2 \rightarrow \mathbb{R}^2$ function is bijective.

Question

Given $f: \mathbb{R}\rightarrow\mathbb{R}$ is a $C^1$ function and $\forall t.|f'(t)| \leq c < 1$,$g: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ is defined as $g(x,y) = (x + f(y), y+f(x)).$ I am trying to show $g$ is bijective.

It is sufficient to show $g$ is 1-to-1 and onto, but how exactly is this approached with such an $\mathbb{R}^2 \rightarrow \mathbb{R}^2$ function?

See the third question in the list of related questions: http://math.stackexchange.com/questions/467322/prove-that-t-mathbbr2-rightarrow-mathbbr2-is-a-bijection?rq=1 — Samuel, Aug 10 '15 at 00:22

score -1 · Answer 1 · answered Aug 10 '15 at 00:11

-1

You can construct the inverse to show the function is bijective: Assuming f bijective:

take $h:= (x-f(y), y-f(x) )$ gives you the inverse, so $g$ is bijective.

answered Aug 10 '15 at 00:11

Gary.

2,432

Sorry, my battery is dying, I will be back in a few hours. – Gary. Aug 10 '15 at 00:15
That's not the inverse. – Ian Aug 10 '15 at 00:15
This is incomplet answer try to give your inverse correctly – zeraoulia rafik Aug 10 '15 at 00:25

Proving $\mathbb{R}^2 \rightarrow \mathbb{R}^2$ function is bijective.

1 Answers1